Fig. 6.
All of the points
of a which lie upon the same side of O, when taken together, are called the
half-ray emanating from O. Hence, each point of a straight line divides
it into two half-rays.
Making use
of the notation of theorem 5, we say: The points A, A0
lie in the plane
upon one and the
same side of the straight line a, and the points A, B lie in the plane upon different
sides of the straight line a.
Definitions. A
system of segments AB, BC, CD, . . . , KL is called a broken line joining A with L
and is designated, briefly, as the broken line ABCDE . . .KL. The
points lying
within the segments AB, BC, CD, . . . , KL, as also the points A, B, C,
D, . . . , K, L, are
called the points of the broken line. In particular, if the point A
coincides with L, the broken
line is called a polygon and is designated as the polygon ABCD. . .K.
The segments AB,
BC, CD, . . . , KA are called the sides of the polygon and the points A, B, C, D, . . .
, K the vertices. Polygons having 3, 4, 5, . . . , n vertices are
called, respectively,
triangles, quadrangles, pentagons, . . . , n-gons. If the vertices of a
polygon are all distinct
and none of them lie within the segments composing the sides of the polygon, and,
furthermore, if no two sides have a point in common, then the polygon is
called a simple
polygon.
With the aid of
theorem 5, we may now obtain, without serious difficulty, the following theorems:
Theorem 6. Every
simple polygon, whose vertices all lie in a plane
, divides the points of this
plane, not belonging to the broken line constituting the sides of the polygon,
into two regions, an interior and an exterior, having the following properties: If A
is a point of the interior region (interior point) and B a point of the exterior
region (exterior point), then any broken line joining A and B must have at least one
point in common with the polygon. If, on the other hand, A, A0
are two points
of the interior and B, B0 two
points of the exterior region, then there are always
broken lines to be found joining A with A0 and
B with B0 without having a point in common
with the polygon. There exist straight lines in the plane
which lie entirely
outside of the given polygon, but there are none which lie entirely
within it.
Theorem 7. Every
plane
divides the remaining points of space into two regions having the
following properties: Every point A of the one region determines with each point B of
the other region a segment AH, within which lies a point of
.
On the other
hand, any two points A, A0 lying
within the same region determine a segment AA0
containing no point of
.
7
Fig. 7.
Making use
of the notation of theorem 7, we may now say: The points A, A0
are situated in space upon one
and the same side of the plane
, and the points A, B are situated in space upon
different sides of the plane
.
Theorem 7 gives us
the most important facts relating to the order of sequence of the elements of space.
These facts are the results, exclusively, of the axioms already
considered, and, hence, no new
space axioms are required in group II.
§ 5. GROUP
III: AXIOM OF PARALLELS. (EUCLID’S AXIOM.)
The introduction
of this axiom simplifies greatly the fundamental principles of geometry and facilitates in
no small degree its development. This axiom may be expressed as follows:
III. In a plane
there can be drawn through any point A, lying outside of a straight line a, one and
only one straight line which does not intersect the line a. This straight line is
called the parallel to a through the given point A.
This statement of
the axiom of parallels contains two assertions. The first of these is that, in the
plane
, there is always a straight line passing through A which does not intersect the
given line a. The second states that only one such line is possible. The
latter of these
statements is the essential one, and it may also be expressed as
follows:
Theorem 8. If two
straight lines a, b of a plane do not meet a third straight line c of the same plane,
then they do not meet each other.
For, if a, b had a
point A in common, there would then exist in the same plane with c two straight lines
a and b each passing through the point A and not meeting the straight line c. This
condition of affairs is, however, contradictory to the second assertion
contained 8 in the axiom of
parallels as originally stated. Conversely, the second part of the axiom
of parallels, in its
original form, follows as a consequence of theorem 8.
The axiom of
parallels is a plane axiom.
§ 6. GROUP
IV. AXIOMS OF CONGRUENCE.
The axioms of this
group define the idea of congruence or displacement. Segments stand in
a certain relation to one another which is described by the word “congruent.”
IV, I. If A,
B are two points on a straight line a, and if A0 is
a point upon the same or
another straight line a0,
then, upon a given side of A0 on
the straight line a0,
we can always find one and only one point B0 so
that the segment AB (or BA) is congruent
to the segment A0B0.
We indicate this relation by writing AB A0B0.
Every segment is
congruent to itself; that is, we always have AB AB.
We can state the
above axiom briefly by saying that every segment can be laid off upon a given side
of a given point of a given straight line in one and and only one way.
IV, 2. If a
segment AB is congruent to the segment A0B0
and also to the segment A00B00,
then the segment A0B0
is congruent to the segment A00B00;
that is, if AB A0B
and AB A00B00,
then A0B0
A00B00.
IV, 3. Let AB and
BC be two segments of a straight line a which have no points in common
aside from the point B, and, furthermore, let A0B0
and B0C0
be two segments of
the same or of another straight line a0 having,
likewise, no point other than B0
in common. Then, if AB A0B0
and BC B0C0,
we have AC A0C0.
Fig. 8.
Definitions. Let
be any arbitrary plane and h, k any two distinct half-rays lying in
and emanating from the point O so as to form a part of two different
straight lines.
We call the system
formed by these two half-rays h, k an angle and represent it by the symbol \(h, k) or
\(k, h). From axioms II, 1–5, it follows readily that the half-rays h
and 9 k, taken together
with the point O, divide the remaining points of the plane a into two regions having the
following property: If A is a point of one region and B a point of the other, then every
broken line joining A and B either passes through O or has a point in common with
one of the half-rays h, k. If, however, A, A0 both
lie within the same region, then it is always
possible to join these two points by a broken line which neither passes through O nor has
a point in common with either of the half-rays h, k. One of these two regions is
distinguished from the other in that the segment joining any two points
of this region lies
entirely within the region. The region so characterised is called the
interior of the angle (h, k).
To distinguish the other region from this, we call it the exterior of
the angle (h, k). The
half rays h and k are called the sides of the angle, and the point O is called the vertex
of the angle.
IV, 4. Let
an angle (h, k) be given in the plane
and let a straight line a0 be
given in a plane 0. Suppose also that, in the
plane
, a definite side of the straight line a0 be assigned.
Denote by h0 a half-ray of
the straight line a0 emanating
from a point O0
of this line. Then in the plane
0 there is one and only one
half-ray k0 such that the angle
(h, k), or (k, h), is congruent to the angle (h0,
k0) and at the same time all interior
points of the angle (h0, k0)
lie upon the given side of a0.
We express this relation by means
of the notation \(h, k)
\(h0, k0)
Every angle is
congruent to itself; that is, \(h, k) \(h,
k) or \(h, k) \(k,
h)
We say, briefly,
that every angle in a given plane can be laid off upon a given side of a given half-ray
in one and only one way.
IV, 5. If
the angle (h, k) is congruent to the angle (h0,
k0) and to the angle (h00,
k00), then the angle (h0,
k0) is congruent to the
angle (h00, k00);
that is to say, if \(h, k) \(h0,
k0) and \(h, k) \(h00,
k00), then \(h0,
k0) \(h00,
k00).
Suppose we have
given a triangle ABC. Denote by h, k the two half-rays emanating from A and passing
respectively through B and C. The angle (h, k) is then said to be the angle included by
the sides AB and AC, or the one opposite to the side BC in the triangle ABC. It contains
all of the interior points of the triangle ABC and is represented by the
symbol \BAC, or by
\A.
IV, 6. If,
in the two triangles ABC and A0B0C0
the congruences AB A0B0,
AC A0C0,
\BAC \B0A0C0
hold, then the
congruences \ABC
\A0B0C0
and \ACB \A0C0B0
also hold.
10
Axioms IV, 1–3
contain statements concerning the congruence of segments of a straight line only. They
may, therefore, be called the linear axioms of group IV. Axioms IV, 4, 5
contain statements
relating to the congruence of angles. Axiom IV, 6 gives the connection between the
congruence of segments and the congruence of angles. Axioms IV, 4–6
contain statements
regarding the elements of plane geometry and may be called the plane
axioms of group IV.
§ 7. CONSEQUENCES
OF THE AXIOMS OF CONGRUENCE.
Suppose the
segment AB is congruent to the segment A0B0.
Since, according to axiom IV, 1, the
segment AB is congruent to itself, it follows from axiom IV, 2 that A0B0
is congruent to
AB; that is to say, if AB A0B0,
then A0B0
AB. We say, then, that the two segments are
congruent to one another.
Let A, B, C,
D, . . . , K, L and A0, B0,
C0, D0,
. . . , K0, L0
be two series of points on the straight
lines a and a0,
respectively, so that all the corresponding segments AB and A0B0,
AC and A0C0,
BC and B0C0,
. . . , KL and K0L0
are respectively congruent, then the two series of
points are said to be congruent to one another. A and A0,
B and B0, . . . , L and L0
are called corresponding points of the two
congruent series of points.
From the linear
axioms IV, 1–3, we can easily deduce the following theorems:
Theorem 9. If the
first of two congruent series of points A, B, C, D, . . . , K, L and A0,
B0, C0,
D0, . . . , K0,
L0 is so arranged that B
lies between A and C, D, . . . , K, L, and C
between A, B and D, . . . , K, L, etc., then the points A0,
B0, C0,
D0, . . . , K0,
L0
of the second series are arranged in a similar
way; that is to say, B0 lies
between A0
and C0,
D0, . . . , K0,
L0, and C0
lies between A0,
B0 and D0,
. . . , K0, L0,
etc. Let the
angle (h, k) be congruent to the angle (h0,
k0). Since, according to
axiom IV, 4, the angle
(h, k) is congruent to itself, it follows from axiom IV, 5 that the
angle (h0, k0)
is congruent to
the angle (h, k). We say, then, that the angles (h, k) and (h0,
k0) are congruent to one another.
Definitions. Two
angles having the same vertex and one side in common, while the sides not
common form a straight line, are called supplementary angles. Two angles
having a common
vertex and whose sides form straight lines are called vertical angles.
An angle which is
congruent to its supplementary angle is called a right angle.
Two
triangles ABC and A0B0C0
are said to be congruent to one another when all
of the following
congruences are fulfilled: AB A0B0,
AC A0C0,
BC B0C0,
\A \A0,
\B \B0, \C \C0.
Theorem 10. (First
theorem of congruence for triangles). If, for the two triangles ABC and A0B0C0,
the congruences AB A0B0,
AC A0C0,
\A \A0 hold, then the two
triangles are congruent to each other.
11
Proof.
From axiom IV, 6, it follows that the two congruences \B \B0
and \C \C0 are
fulfilled, and it is, therefore, sufficient to show that the two sides
BC and B0C0
are congruent.
We will assume the contrary to be true, namely, that BC and B0C0
are not congruent,
and show that this leads to a contradiction. We take upon B0C0
a point D0 such that BC
B0D0.
The two triangles ABC and A0B0D0
have, then, two sides and the included angle
of the one agreeing, respectively, to two sides and the included angle of the
other. It follows from axiom IV, 6 that the two angles BAC and B0A0D0
are also congruent to
each other. Consequently, by aid of axiom IV, 5, the two angles B0A0C0
and B0A0D0
must be congruent.
Fig. 9.
This, however, is
impossible, since, by axiom IV, 4, an angle can be laid off in one and only one way on a
given side of a given half-ray of a plane. From this contradiction the theorem follows.
We can also easily
demonstrate the following theorem:
Theorem 11.
(Second theorem of congruence for triangles). If in any two triangles one side and the
two adjacent angles are respectively congruent, the triangles are congruent.
We are now in a
position to demonstrate the following important proposition.
Theorem 12.
If two angles ABC and A0B0C0
are congruent to each other, their supplementary
angles CBD and C0B0D0
are also congruent.
Fig. 10.
Proof. Take
the points A0, C0,
D0 upon the sides passing
through B0 in such a way that
A0B0
AB, C0B0
CB, D0B0
DB.
Then, in the
two triangles ABC and A0B0C0,
the sides AB and BC are respectively congruent to
A0B0 and
C0B0.
Moreover, since the angles included by these sides are 12
congruent to each
other by hypothesis, it follows from theorem 10 that these triangles are
congruent; that is
to say, we have the congruences AC A0C,
\BAC \B0A0C0.
On the other
hand, since by axiom IV, 3 the segments AD and A0D0
are congruent to each other, it
follows again from theorem 10 that the triangles CAD and C0A0D0
are congruent, and, consequently,
we have the congruences:
CD C0D0,
\ADC \A0D0C0.
From these
congruences and the consideration of the triangles BCD and B0C0D0,
it follows by virtue of
axiom IV, 6 that the angles CBD and C0B0D0
are congruent.
As an immediate
consequence of theorem 12, we have a similar theorem concerning the congruence of
vertical angles.
Theorem 13.
Let the angle (h, k) of the plane
be congruent to the angle (h0,
k0) of the plane 0, and, furthermore, let l
be a half-ray in the plane
emanating from the vertex of the
angle (h, k) and lying within this angle. Then, there always exists in the plane
0 a half-ray l0
emanating from the vertex of the angle (h0,
k0) and lying within this angle
so that we have \(h, l)
\(h0, l0),
\(k, l) \(k0, l0).
Fig. 11.
Proof. We
will represent the vertices of the angles (h, k) and (h0,
k0) by O and O0,
respectively,
and so select upon the sides h, k, h0,
k0 the points A, B, A0,
B0 that the congruences
OA O0A0,
OB O0B0
are
fulfilled. Because of the congruence of the triangles OAB and O0A0B0,
we have at once AB A0B0,
\OAB O0A0B0,
\OBA \O0B0A0.
Let the
straight line AB intersect l in C. Take the point C0
upon the segment A0B0
so that A0C0
AC. Then, O0C0
is the required half-ray. In fact, it follows
directly from these congruences, by aid of axiom IV, 3, that BC B0C0.
Furthermore, the triangles OAC and O0A0C0
are congruent to each other, and the same is true
also of the triangles OCB and O0B0C0.
With this our proposition is demonstrated.
In a similar
manner, we obtain the following proposition.
13
Theorem 14.
Let h, k, l and h0, k0,
l0 be two sets of three
half-rays, where those of each set
emanate from the same point and lie in the same plane. Then, if the congruences
\(h, l)
\(h0, l0),
\(k, l) \(k0, l0)
are fulfilled, the
following congruence is also valid; viz.: \(h, k)
\(h0, k0).
By aid of theorems
12 and 13, it is possible to deduce the following simple theorem, which Euclid held–although
it seems to me wrongly–to be an axiom.
Theorem 15. All
right angles are congruent to one another. Proof. Let the
angle BAD be congruent to its supplementary angle CAD, and, likewise,
let the angle B0A0D0
be congruent to its supplementary angle C0A0D0.
Hence the angles BAD,
CAD, B0A0D0,
and C0A0D0
are all right angles. We will assume that the contrary of
our proposition is true, namely, that the right angle B0A0D0
is not congruent to the right angle
BAD, and will show that this assumption leads to a contradiction. We lay off the
angle B0A0D0
upon the half-ray AB in such a manner that the
side AD00 arising from this
operation falls either within the angle BAD or within the angle CAD.
Suppose, for example, the
first of these possibilities to be true. Because of the congruence of
the angles B0A0D0
and BAD00,
it follows from theorem 12 that angle C0A0D0
is congruent to angle CAD00,
and, as the angles B0A0D0
and C0A0D0
are congruent to each other, then, by IV, 5, the
angle BAD00 must be
congruent to CAD00.
Fig. 12.
Furthermore, since
the angle BAD is congruent to the angle CAD, it is possible, by theorem 13,
to find within the angle CAD a half-ray AD000 emanating
from A, so that the angle
BAD00 will be congruent to
the angle CAD000, and also
the angle DAD00 will be congruent to
the angle DAD000. The angle
BAD00 was shown to be
congruent to the angle CAD00
and, hence, by axiom IV, 5, the angle CAD00,
is congruent to the angle CAD000.
This, however, is
not possible; for, according to axiom IV, 4, an angle can be laid off in
a plane upon a
given side of a given half-ray in only one way. With this our
proposition is demonstrated.
We can now introduce, in accordance with common usage, the terms “acute
angle” and “obtuse angle.”
14
The theorem
relating to the congruence of the base angles A and B of an equilateral triangle ABC
follows immediately by the application of axiom IV, 6 to the triangles
ABC and BAC. By aid of
this theorem, in addition to theorem 14, we can easily demonstrate the following
proposition.
Theorem 16. (Third
theorem of congruence for triangles.) If two triangles have the three sides of
one congruent respectively to the corresponding three sides of the other, the
triangles are congruent. Any finite number
of points is called a figure. If all of the points lie in a plane, the figure is called a
plane figure. Two figures are
said to be congruent if their points can be arranged in a one-to-one correspondence so
that the corresponding segments and the corresponding angles of the two figures are in
every case congruent to each other.
Congruent figures
have, as may be seen from theorems 9 and 12, the following properties:
Three points of a
figure lying in a straight line are likewise in a straight line in every
figure congruent
to it. In congruent figures, the arrangement of the points in
corresponding planes with
respect to corresponding lines is always the same. The same is true of
the sequence of
corresponding points situated on corresponding lines. The most general
theorems relating to congruences in a plane and in space may be expressed as
follows:
Theorem
17.
If (A,B,C, . . .) and (A0,B0,C0,
. . .) are congruent plane figures and P is a point in
the plane of the first, then it is always possible to find a point P in the plane of
the second figure so that (A,B,C, . . . , P) and (A0,B0,C0,
. . . , P0) shall likewise be
congruent figures. If the two figures have at least three points not
lying in a
straight line, then the selection of P0 can
be made in only one way.
Theorem
18.
If (A,B,C, . . .) and (A0,B0,C0,
. . . = are congruent figures and P represents
any arbitrary point, then there can always be found a point P0
so that the two
figures (A,B,C, . . . , P) and (A0,B0,C0,
. . . , P0) shall likewise
be congruent. If the figure (A,B,C,
. . . , P) contains at least four points not lying in the same plane, then the
determination of P0 can be
made in but one way.
This theorem
contains an important result; namely, that all the facts concerning
space which have
reference to congruence, that is to say, to displacements in space, are
(by the addition of the
axioms of groups I and II) exclusively the consequences of the six
linear and plane axioms
mentioned above. Hence, it is not necessary to assume the axiom of parallels in order
to establish these facts.
If we take, in,
addition to the axioms of congruence, the axiom of parallels, we can then easily
establish the following propositions:
Theorem 19. If two parallel lines are cut by a third straight line, the
alternateinterior angles and also
the exterior-interior angles are congruent Conversely, if the alternate-interior
or the exterior-interior angles are congruent, the given lines are parallel.
15
Theorem 20. The
sum of the angles of a triangle is two right angles.
Definitions. If M
is an arbitrary point in the plane
, the totality of all points A, for which the
segments MA are congruent to one another, is called a circle. M is
called the centre of the
circle. From this
definition can be easily deduced, with the help of the axioms of groups
III and IV, the known
properties of the circle; in particular, the possibility of constructing
a circle through any
three points not lying in a straight line, as also the congruence of all
angles inscribed
in the same segment of a circle, and the theorem relating to the angles
of an inscribed
quadrilateral.
§ 8. GROUP
V. AXIOM OF CONTINUITY. (ARCHIMEDEAN AXIOM.)
This axiom makes
possible the introduction into geometry of the idea of continuity. In order to state
this axiom, we must first establish a convention concerning the equality
of two segments.
For this purpose, we can either base our idea of equality upon the axioms relating to
the congruence of segments and define as “equal” the correspondingly
congruent
segments, or upon the basis of groups I and II, we may determine how, by
suitable
constructions (see Chap. V, § 24), a segment is to be laid off from a
point of a given straight
line so that a new, definite segment is obtained “equal” to it. In
conformity with such a
convention, the axiom of Archimedes may be stated as follows:
V. Let A1
be any point upon a straight line between the
arbitrarily chosen points A and B.
Take the points A2, A3,
A4, . . . so that A1
lies between A and A2,
A2
between A1 and
A3, A3
between A2 and
A4 etc. Moreover, let the
segments AA1,
A1A2,
A2A3,
A3A4,
. . . be equal to one
another. Then, among this series of points, there always exists a certain
point An such that B lies
between A and An.
The axiom of
Archimedes is a linear axiom. Remark.3
To the preceeding five groups of axioms, we may
add the following one, which, although
not of a purely geometrical nature, merits particular attention from a theoretical point
of view. It may be expressed in the following form:
Axiom of
Completeness.4 (Vollst¨andigkeit):
To a system of points, straight lines, and planes,
it is impossible to add other elements in such a manner that the system thus
generalized shall form a new geometry obeying all of the five groups of axioms.
In other words, the elements of geometry form a system which is not
susceptible of extension, if we regard the five groups of axioms as valid.
3Added by
Professor Hilbert in the French translation.—Tr.
4See Hilbert, “Ueber
den Zahlenbegriff,” Berichte der deutschen Mathematiker-Vereinigung,
1900.
16
This axiom gives
us nothing directly concerning the existence of limiting points, or of the idea of
convergence. Nevertheless, it enables us to demonstrate Bolzano’s
theorem by virtue of which,
for all sets of points situated upon a straight line between two
definite points of the same
line, there exists necessarily a point of condensation, that is to say, a limiting point.
From a theoretical point of view, the value of this axiom is that it leads indirectly
to the introduction of limiting points, and, hence, renders it possible
to establish a
one-to-one correspondence between the points of a segment and the system
of real numbers.
However, in what is to follow, no use will be made of the “axiom of completeness.”
17
COMPATIBILITY AND
MUTUAL INDEPENDENCE OF THE
AXIOMS.
§ 9. COMPATIBILITY
OF THE AXIOMS.
The axioms, which
we have discussed in the previous chapter and have divided into five groups, are not
contradictory to one another; that is to say, it is not possible to
deduce from these axioms,
by any logical process of reasoning, a proposition which is
contradictory to any of them. To
demonstrate this, it is sufficient to construct a geometry where all of the five groups
are fulfilled. To this end, let
us consider a domain consisting of all of those algebraic numbers which may be
obtained by beginning with the number one and applying to it a finite number of times
the four arithmetical operations (addition, subtraction, multiplication,
and
division) and the operation p1 + !2,
where ! represents a number arising from the five operations
already given.
Let us regard a
pair of numbers (x, y) of the domain as defining a point and the ratio of three
such numbers (u : v : w) of , where u, v are not both equal to zero, as defining a
straight line. Furthermore, let the existence of the equation ux
+ vy + w = 0 express the
condition that the point (x, y) lies on the straight line (u : v : w).
Then, as one readily sees,
axioms I, 1–2 and III are fulfilled. The numbers of the domain are all
real numbers. If now we
take into consideration the fact that these numbers may be arranged according to
magnitude, we can easily make such necessary conventions concerning our points and
straight lines as will also make the axioms of order (group II) hold. In
fact, if (x1,
y1), (x2,
y2), (x3,
y3), . . . are any points
whatever of a straight line, then this may be taken as
their sequence on this straight line, providing the numbers x1,
x2, x3,
. . . , or the numbers
y1, y2,
y3, . . . , either all
increase or decrease in the order of sequence given here. In order
that axiom II, 5 shall be fulfilled, we have merely to assume that all
points corresponding to
values of x and y which make ux + vy + w less than zero or greater than zero shall
fall respectively upon the one side or upon the other side of the
straight line (u : v : w).
We can easily convince ourselves that this convention is in accordance with those which
precede, and by which the sequence of the points on a straight line has already been
determined.
The laying off of
segments and of angles follows by the known methods of analytical geometry. A
transformation of the form
x0
= x + a
y0
= y + b
produces a
translation of segments and of angles.
18
Fig. 13.
Furthermore, if,
in the accompanying figure, we represent the point (0, 0) by O and the point (1, 0)
by E, then, corresponding to a rotation of the angle COE about O as a center, any
point (x, y) is transformed into another point (x0,
y0) so related that
x0
= a
pa2
+ b2
x −
b
pa2
+ b2
y,
y0
= b
pa2
+ b2
x + a
pa2
+ b2
y.
Since the number
pa2
+ b2 =
as1 + b
a2
belongs to the
domain , it follows that, under the conventions which we have made, the axioms of
congruence (group IV) are all fulfilled. The same is true of the axiom
of Archimedes.
Fig. 14.
From these
considerations, it follows that every contradiction resulting from our
system of axioms must
also appear in the arithmetic related to the domain . The corresponding
considerations for the geometry of space present no difficulties. If, in the
preceding development, we had selected the domain of all real numbers instead of the
domain , we should have obtained likewise a geometry in which all of the
axioms of groups I—V
are valid. For the purposes of our demonstration, however, it was sufficient to take
the domain , containing on an enumerable set of elements.
19
§ 10. INDEPENDENCE
OF THE AXIOMS OF PARALLELS.
(NON-EUCLIDEAN
GEOMETRY.)5
Having shown that
the axioms of the above system are not contradictory to one another, it is of interest
to investigate the question of their mutual independence. In fact, it
may be shown that none of
them can be deduced from the remaining ones by any logical process of reasoning.
First of all, so
far as the particular axioms of groups I, II, and IV are concerned, it is easy to show
that the axioms of these groups are each independent of the other of the
same group.6
According to our
presentation, the axioms of groups I and II form the basis of the remaining axioms.
It is sufficient, therefore, to show that each of the groups II, IV, and
V is independent
of the others.
The first
statement of the axiom of parallels can be demonstrated by aid of the
axioms of groups I, II,
and IV. In order to do this, join the given point A with any arbitrary
point B of the straight
line a. Let C be any other point of the given straight line. At the
point A on AB, construct
the angle ABC so that it shall lie in the same plane
as the point C, but upon the
opposite side of AB from it. The straight line thus obtained through A does not meet the
given straight line a; for, if it should cut it, say in the point D, and
if we
suppose B to be situated between C and D, we could then find on a a
point D0 so situated
that B would lie between D and D0,
and, moreover, so that we should have AD BD0
Because of
the congruence of the two triangles ABD and BAD0,
we have also \ABD
\BAD0, and since
the angles ABD0 and ABD are
supplementary, it follows from theorem 12 that the angles
BAD and BAD0 are also
supplementary. This, however, cannot be true, as, by theorem 1, two
straight lines cannot intersect in more than one point, which would be the case if
BAD and BAD0 were
supplementary.
The second
statement of the axiom of parallels is independent of all the other
axioms. This may be most
easily shown in the following well known manner. As the individual elements of a
geometry of space, select the points, straight lines, and planes of the
ordinary geometry as
constructed in § 9, and regard these elements as restricted in extent
to the interior of a
fixed sphere. Then, define the congruences of this geometry by aid of
such linear
transformations of the ordinary geometry as transform the fixed sphere
into itself.
By suitable
conventions, we can make this “non-euclidean geometry” obey all of
the axioms of our system
except the axiom of Euclid (group III). Since the possibility of the
ordinary geometry has
already been established, that of the non-euclidean geometry is now an immediate
consequence of the above considerations.
5The mutual
independence of Hilbert’s system of axioms has also been discussed
recently by Schur and Moore. Schur’s
paper, entitled “Ueber die Grundlagen der Geometrie” appeared in
Math. Annalem, Vol. 55, p. 265, and
that of Moore, “On the Projective Axioms of Geometry,” is to be
found in the Jan. (1902) number of the
Transactions of the Amer. Math. Society.—Tr. 6See my lectures
upon Euclidean Geometry, winter semester of 1898–1899, which were
reported by Dr. Von Schaper and
manifolded for the members of the class.
20
§ 11. INDEPENDENCE
OF THE AXIOMS OF CONGRUENCE.
We shall show the
independence of the axioms of congruence by demonstrating that axiom IV, 6, or what
amounts to the same thing, that the first theorem of congruence for
triangles (theorem 10)
cannot be deduced from the remaining axioms I, II, III, IV 1–5, V by
any logical process of
reasoning.
Select, as the
points, straight lines, and planes of our new geometry of space, the points, straight
lines, and planes of ordinary geometry, and define the laying off of an angle as in
ordinary geometry, for example, as explained in § 9. We will, however,
define the laying
off of segments in another manner. Let A1,
A2 be two points which, in
ordinary geometry,
have the co-ordinates x1, y1,
z1 and x2,
y2, z2,
respectively. We will now define the length
of the segment A1A2
as the positive value of the expression
p(x1
− x2 +
y1 − y2)2
+ (y1 −
y2)2 +
(z1 − z2)2
and call the
two segments A1A2
and A01A02
congruent when they have equal lengths in the sense just
defined. It is at once
evident that, in the geometry of space thus defined, the axioms I, II,
III, IV 1–2, 4–5, V
are all fulfilled.
In order to show
that axiom IV, 3 also holds, we select an arbitrary straight line a and upon it
three points A1, A2,
A3 so that A2
shall lie between A1 and
A3. Let the points x, y, z of the
straight line a be given by means of the equations
x = t +
0,
y = μt
+ μ0,
z = t +
0,
where ,
0, μ, μ0,
, 0 represent
certain constants and T is a parameter. If t1,
t2 (< t1),
t3
(< t2)
are the values of the parameter corresponding to the points A1,
A2, A3
we have as the
lengths of the three segments A1A2
A2A3
and A1A3
respectively, the following values:
(t1
− t2)
p( + μ)2 +
μ2 + 2
(t2 − t3)
p( + μ)2 +
μ2 + 2
(t1 − t3)
p( + μ)2 +
μ2 + 2
Consequently, the length of A1A3
is equal to the sum of the lengths of the segments
A1A2 and A2A3.
But this result is equivalent to the existence of axiom IV, 3.
Axiom IV, 6, or
rather the first theorem of congruence for triangles, is not always fulfilled in this
geometry. Consider, for example, in the plane z = 0, the four points O, having the
co-ordinates x = 0, y = 0
A, “ “ “ x =
1, y = 0
B, “ “ “ x =
0, y = 1
C, “ “
“ x = 1
2 ,
y = 1
2
21
Fig. 15.
Then, in the right
triangles OAC and OBC, the angles at C as also the adjacent sides AC and BC are
respectively congruent; for, the side OC is common to the two triangles
and the sides AC
and BC have the same length, namely, 1 2 .
However, the third sides OA and OB have the
lengths 1 and p2, respectively, and are not, therefore, congruent. It is
not difficult to find
in this geometry two triangles for which axiom IV, 6, itself is not
valid.
§ 12. INDEPENDENCE
OF THE AXIOM OF CONTINUITY.
(NO
